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3 years ago

Which one of the following pairs of separation techniques will BOTH separate salt from a mixture of salt and water?

Chemistry

1 answer:

Natali5045456 [20]3 years ago

6 0

Answer:

What are the options and i dont know what the options are but if one of them is distilation im 99.9% sure that is the answer.

Explanation:

Be sure to friend me and mark me as braniliest if im correct!

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The missing components in the table to the right

Fed [463]

The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...

2 answers

4 0

3 years ago

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o

jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

Step 1: The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

Step 2: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

Step 3: Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

Step 4: Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

Step 5: Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

Step 6: Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

7 0

3 years ago

A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K

Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K

The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0

3 years ago

Read 2 more answers

The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of

Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be

10 g

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of

1 g

of that substance by

∘

Water has a specific heat of approximately

4.18

∘

. This tells you that in order to increase the temperature of

1 g

of water by

∘

, you need to provide

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of

1 g

of water by

∘

Well, you'd need

4.18 J

to increase it by

∘

, another

4.18 J

to increase it by another

∘

, and so on. This means that you'd need

4.18 J

41.8 J

to increase the temperature of

1 g

of water by

∘

Now look at the value given to you. If you need

41.8 J

to increase the temperature of

1 g

of water by

∘

, what mass of water would require

times as much heat to increase its temperature by

∘

1 g

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

⋅

, where

- heat absorbed/lost

- the mass of the sample

- the specific heat of the substance

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

⋅

4.18

∘

⋅

(

−

)

∘

418

4.18

⋅

10 g

5 0

3 years ago

An airplane starts at rest and Excelerator down the runway for 20 seconds. At the end of the runway, its velocity is 80 m/s Nort

masha68 [24]

According to the formula you have given us to work with . . .

1). The airplane's acceleration is

(80 m/s north - zero) / (20 sec) = 4 m/sec^2 north

2). For the cyclist:

(V-final - zero) / 20sec = 0.5 m/s^2 south

Multiply each side by 20s : V-final = 0.5 m/s^2 south x (20sec) =

10 m/s south

8 0

3 years ago